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Let $\mathbb{F}_Q$ be an extension field of $\mathbb{F}_q$. Show that if an element $a\in \mathbb{F}_Q$ is a root of a polynomial $h(x)\in\mathbb{F}_q[x]$, then so are the elements $a^{q^r}$ for $r\geq0$.

First since $h(x)\in\mathbb{F}_q[x]$ and $\mathbb{F}_q$ is a subfield of $\mathbb{F}_Q$ we have $h(x)\in\mathbb{F}_Q[x]$. For $r=0$ it evidently holds. Can you give some ideas for $r>0$?

EDIT Based on Stahl's suggestions I have proceeded to the following:

Let $h(x)=\sum\limits_{i=0}^mb_ix^i$ for $b_i\in\mathbb{F}_q$, $\mathrm{degree}(h(x))=m$.

Now $a^q\in\mathbb{F}_Q$ since $a\in\mathbb{F}_Q$. Then

\begin{eqnarray} h(a^q)&=&\sum\limits_{i=0}^mb_ia^{qi}\\ &=&\sum\limits_{i=0}^mb_i^qa^{qi} \quad\text{since $b_i^q=b_i$ for } b_i\in\mathbb{F}_q\\ &=&\sum\limits_{i=0}^m(b_ia^{i})^q\\ &&\text{by property $(a+b)^{p^n}=a^{p^n}+b^{p^n}$ and the order of the field can be expressed as some}\\ &&\text{power of its characteristic $q=p^n$}\\ &=&\left(\sum\limits_{i=0}^mb_ia^{i}\right)^q\\ &=&(h(a))^q\\ &=&0 \end{eqnarray}

So for $r=1$ it holds (induction basis). Assume it holds for $r=n$ i.e. $h(a^{q^n})=0$ for $n>1$ (induction hypothesis). For $r=n+1$: $$h(a^{q^{n+1}})=\sum\limits_{i=0}^mb_ia^{q^{n+1}i}=\sum\limits_{i=0}^mb_i^q(a^{q^{n}i})^q=\sum\limits_{i=0}^m(b_ia^{q^{n}i})^q=\left(\sum\limits_{i=0}^mb_ia^{q^{n}i}\right)^q=0$$ verifying the induction step.

mgus
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    Looks good to me. It may be slightly cleaner to prove by the same recipe that $h(z^q)=h(z)^q$ for any $z$ in any extension field of $\Bbb{F}_q$ (you actually do just that!). Then you only need to do that calculation once. Applying that to $z=a$ gives you the base case whereas applying it to $z=a^{q^n}$ gives the inductive step. Not much difference though. – Jyrki Lahtonen Oct 01 '17 at 06:26
  • The same theme has been covered on our site earlier: 1, 2, and possibly earlier and better explanations. Since I tend to be involved with questions in this tag I should not cast a vote to close as a dupe. – Jyrki Lahtonen Oct 01 '17 at 07:10
  • This may actually be the most informative earlier incarnation. – Jyrki Lahtonen Oct 01 '17 at 07:11
  • @JyrkiLahtonen Thanks for pointing that out. I always search for previous posts before posting but with mixed results of success. – mgus Oct 01 '17 at 07:14
  • In that case, mgus, you did more than most askers! As the internal search engine won't grok LaTeX using it is a bit taxing. A good faith attempt is enough. I have a bit more experience feeding it with nice buzzwords. – Jyrki Lahtonen Oct 01 '17 at 07:17
  • I find this question confusing since the title asserts that $a \in \Bbb F_Q$ satisfies $h(x) \in \Bbb F_q[x]$, but the first sentence in the body of the question text says $a \in \Bbb F_q$ satisfies $h(x) \in \Bbb F_Q[x]$, thus reversing the roles of $\Bbb F_q$ and $\Bbb F_Q$ as indicated in the title. – Robert Lewis Oct 01 '17 at 17:11
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    @RobertLewis I fixed it. Thank you. – mgus Oct 01 '17 at 18:25
  • Are you sure you don't mean $a \in \Bbb F_Q$ in the first sentence of the text of the question? – Robert Lewis Oct 01 '17 at 18:31
  • @RobertLewis Uhh.. A lot of typos. Thanks. Can you check? – mgus Oct 01 '17 at 19:00
  • The question looks OK now, but I haven't checked the details of the part based on Stahl's answer. – Robert Lewis Oct 01 '17 at 19:08

1 Answers1

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Hint 1: In characteristic $p,$ $(a + b)^p = a^p + b^p.$ Use Lagrange's theorem on $\Bbb F_q^\times$ for $r = 1$ and finish the proof via induction on $r.$

Hint 2: Raise both sides of the equation $h(a) = 0$ to the $q^r$-th power and use the above hint.

Alternate Hint: $\operatorname{Gal}(\Bbb F_{q^n}/\Bbb F_q)$ is generated by $\sigma : x\mapsto x^q.$

Stahl
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  • It doesn't help much. Can we claim that the characteristic of the field here is $q=p^m$ – mgus Oct 01 '17 at 04:17
  • @mgus If $q = p^m,$ then the characteristic of $\Bbb F_q$ is $p;$ every finite field has prime characteristic. If you aren't aware of this, you should spend a minute proving it (it's not very difficult). – Stahl Oct 01 '17 at 04:20
  • Yeap, I know the proof is by assuming it composite. It was my typo. Anyway, can you provide more help to the problem? – mgus Oct 01 '17 at 04:31
  • @mgus I've added another hint, see if you can figure out the proof now. – Stahl Oct 01 '17 at 04:40
  • I have made some progress based on your suggestions but I am not sure about the induction. Could you check the edit? – mgus Oct 01 '17 at 06:13